Quantum String action

Question

If we have an action of the form:

$$ S^{(0)} =\int d^Dx\sqrt{-G}\left(R+\alpha (\partial \phi)^2 + \frac{1}{6} e^{2 \alpha \phi}H_{mnp}H^{mnp}\right).$$

where $H_{mnp}=\partial_m B_{np}+\partial_n B_{pm}+\partial_p B_{mn}$ , $B_{mn}$ is an anti-symmetric tensor and $\alpha= \frac{4}{D-2}$ . This is the effective action of lowest order in the, so called, s parametrization in the section of massless particles in the theory of quantum strings. In order to find the equations for the anti-symetric tensor I did the following:

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\delta\left(R+\alpha (\partial \phi)^2 + \frac{1}{6} e^{2 \alpha \phi}H_{mnp}H^{mnp}\right) $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{6} e^{2 \alpha \phi} \delta (H_{mnp}H^{mnp})\right) $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \delta (H_{mnp})H^{mnp}\right) $$

Now:

$$\delta (H_{mnp})H^{mnp} =\partial_m\delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}] $$

$$ \delta S^{(0)} =\int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \partial_m\delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}] \right) $$

From here I integrated by parts:

$$ \delta S^{(0)} =\int d^{D-1}x\sqrt{F}\left( \frac{1}{3} e^{2 \alpha \phi} \delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] \right) \Big|_a^b \ - \int d^Dx\sqrt{-G}\left( \frac{1}{3} e^{2 \alpha \phi} \delta(B_{np})[H^{mnp}+H^{pmn}+H^{npm}]_{,m} \right) $$

and the first integral has its limits of integration that I labelled as $a$ and $b$. Do I get rid of the first term as it´s being evaluated in the respective limits (something similar to what we do in the anti symmetric Faraday tensor) and the 2nd term yields me the equation Im looking for? Or is my procedure wrong?

Edit:

It´s most likely wrong because the metric is D-dimentional but I dont know how to do it any other way.

Edit 2:

The equations I got for the metric field:

$$ -\frac{1}{2}G_{\mu\nu}[R+ \alpha (\partial \phi)^2+\frac{1}{6} e^{2 \alpha \phi} H_{mnp}H^{mnp}]+R_{\mu\nu}+\alpha \partial_\mu \phi \partial_\nu \phi + \frac{1}{2} e^{2 \alpha \phi} H_{\mu np}H_{\nu}^{\space np} = 0$$

And for the field $\phi$:

$$ - \frac{1}{3} e^{2 \alpha \phi} H_{mnp}H^{mnp} + 2 \alpha \square \phi = 0$$

Edit 3:

The equation I got for the anti-symmetric tensor:

$$ \partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0 $$

Answer 1

The variation of the field $B$ vanishes at the boundary thus the total derivative term vanishes. Now regarding the last term, the derivative should also act on the exponential. When you apply Leibniz rule you have:

$$\partial(e^{2 \alpha \phi}\delta(B_{np}) [H^{mnp}+H^{pmn}+H^{npm}] )=0 $$

as a total divergence term. The derivative acts on three terms, you forgot the exponential.

Answer 2

$$\partial_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$

Shouldn't be?

$$\nabla_m [e^{2 \alpha \phi} (H^{mnp}+H^{pmn}+H^{npm})] = 0$$